Answer
The arc length is proportional to $a$. See proof below.
Work Step by Step
Using implicit differentiation, we find $y^{\prime}=-(x / y)^{r-1}$ and
\begin{align*}
1+\left(y^{\prime}\right)^{2}&=1+(x / y)^{2 r-2}\\
&=\frac{x^{2 r-1}+y^{2 r-2}}{y^{2 r-2}}\\
&=\frac{x^{2 r-2}+\left(a^{r}-x^{r}\right)^{2-2 / r}}{\left(a^{r}-x^{r}\right)^{2-2 / r}}
\end{align*}
Then $$s=\int_{0}^{a} \sqrt{\frac{x^{2 r-2}+\left(a^{r}-x^{r}\right)^{2-2 / r}}{\left(a^{r}-x^{r}\right)^{2-2 / r}}} d x$$Using the substitution $x=a u,$ we obtain
$$
s=a \int_{0}^{1} \sqrt{\frac{u^{2 r-2}+\left(1-u^{r}\right)^{2-2 / r}}{\left(1-u^{r}\right)^{2-2 / r}}} d u
$$
where the integral is independent of $a$.
Thus, we see that the arc length is proportional to $a$.
